This question is incomplete, the complete question is;
Steam heated at constant pressure in a steam generator enters the first stage of a supercritical reheat cycle at 28 MPa, 520°C. Steam exiting the first-stage turbine at 6 MPa is reheated at constant pressure to 500°C. Each turbine stage has an isentropic efficiency of 78% while the pump has an isentropic efficiency of 82%. Saturated liquid exits the condenser that operates at constant pressure of 6 kPa.
Determine the quality of the steam exiting the second stage of the turbine and the thermal efficiency.
Answer:
- the quality of the steam exiting the second stage of the turbine is 0.9329
- the thermal efficiency is 36.05%
Explanation:
get the properties of steam at pressure p1 = 28 MPa and temperature T2 = 520°C .
Specific enthalpy h1= 3192.3 kJ/kg
Specific entropy s1 = 5.9566 kJ/kg.K
Process 1 to 2s is isentropic expansion process in the turbine
S1 = S2s
get the enthalpy at state 2s at pressure p2 = 6 MPa and S2s = 5.9566 kJ/kg.K
h2s = 2822.2 kJ/kg
get the enthalpy at state 2 using isentropic turbine efficiency of the turbine. nT1 = (h1 - h2) / (h1 - h2s)
0.78 = (3192.3 - h2) / (3192.3 - 2822.2)
h2 = 2903.6 kJ/kg
get the enthalpy at state 3 at pressure p2 = p3 = 6 MPa and T3 = 500°C
h3 = 3422.2 kJ/kg
s3 = 6.8803 kJ/kg.K
Process 3 to 4s is isentropic expansion process in the turbine
S3 = S4s
get the enthalpy at state 4s at pressure p4s = p4 = 6 kPa and S4s = 6.8803 kJ/kg.K
h4s = 2118.8 kJ/kg
get the enthalpy at state 4 using isentropic turbine efficiency of the turbine. nT2 = (h3 - h4) / (h3 - h4s)
0.78 = (3422.2 - h4) / ( 3422.2 - 2118.8 )
h4 = 2405.5 kJ/kg
get the properties at pressure, p5 = 6 kPa
h5 = hf
= 151.53 kJ/kg
v5 = Vf
= 0.0010064 m³/kg
get the enthalpy at state 6 using isentropic pump efficiency of the turbine, at
p6 = p1 = 28 MPa
np = v5( p6 - p5) / (h6 - h5)
0.82 = ((0.0010064)( 28000 - 6)) / (h6 - 151.53)
h6 = 185.89 kJ/kg
Now to find the quality of the steam at the exit of the second stage of the turbine
At stat4, p4 = 6kPa
h4f = 151.53 kJ/kg
h4fg = 2415.9 kJ/kg
h4 = h4f + x4h4fg
2405.5 = 151.53 + (x4 (2415.9))
x4 = 0.9329
the quality of the steam exiting the second stage of the turbine is 0.9329
Also to find the efficiency of the power plant, we use the following equation;
n = Wnet / Qin
= (Wt1 + Wt2 - Wp) / (Q61 + Q23)
= [(h1 - h2) + (h3 - h4) - (h6 - h5)] / [(h1 - h6) + (h3 - h2)]
[(3192.3 - 2903.6) + (3422.2 - 2405.5) - (185.89 - 151.53)] / [(3192.3 - 185.89) + (3422.2 - 2903.6)]
= 0.3605
n = 36.05%
therefore the thermal efficiency is 36.05%
